Recent content by Ronaldo95163

Homework Statement I'm just required to setup the integral for the question posted below Homework EquationsThe Attempt at a Solution So solving for phi @ the intersection of the sphere and the plane z=2: z = pcos(phi) 2 = 3cos(phi) phi = arccos(2/3) so my limits for phi would go from 0 to...

oops made a sign error Im getting cos(phi) +/- sqrt(1+cos^2(x) ) And btw is it wrong to use geometry to get rho and phi in certain cases? For example this qn(Not switching the topic of the thread btw...just found a question where I think geometry may be applicable)

Sorry about that...didn't get to respond because of assignments etc for other courses. But I worked that out already but I didnt post it so here it is: x(sq)+y(sq) + z(sq)-2az+a(sq) = 2a(sq) rho(sq)-2az = a(sq) rho(sq)-2a(rho)(cos(phi)) = a(sq) rho(sq)-2a(rho)(cos(phi)) -a(sq) = 0 Using the...

Woa haha yeah I did sorry about that

This was the diagram I used to get phi from the yz plane. I just did cos(phi) which is 2a/rho and found rho from that to get 2asec(phi). But I understand where you're coming from...if I change the eqns I get z = 2r and z=sqrt(2a^2-r)+a. But for spherical...I can't use the diagram like that to...

Ohh I see. From there I graphed the region onto the yz plane. So its the graph z= 2|y| intersected by the sphere. I then used a bit of trigs to get that cos(phi) = 2a/rho giving rho as 2asec(phi) and tan(phi) as 1/2 -> phi = arctan(1/2)So my bounds were: 0<rho<2asec(phi) 0<phi<arctan(1/2)

Well I decided to use spherical coordinates from the start and normally to get my bounds for rho and phi I set x=0 to get the projection on the yz plane. So from the cone's equation I ended up with z=2y and I substituted z=2a into that which gave y=a.

Hmm okay so I solved for z already which I got as 2a ---> Resulting in y = a. Substituting those values back into the equation for the cone I get x = 0. So the curves intersect in the yz plane? But if you substitute z=2a in that same eqn for the cone you get a circle of radius a. So the region...

lol I thought it would have been easier to understand straight from the textbook. But it said z>=0 so at 0 wouldn't the solid be lying on the x-y plane? Because if I substitute z= 0 into the equation of the sphere then you get a circle

^Sorry About that...I've fixed it

Thanks a lot man! What did you use to plot that btw?

Hey guys I've been working some triple integration problems and I've stumbled across a question that I'm having problems with So from the picture below my solution is incorrect and I can't seem to figure out where I went wrong. Is my setup for the integrals correct or is that where I've made my...

Ahh yes they do...the parabola closes down on the xy plane. That means I will be integrating f(x,y) over the region bounded by the two sine functions and the lines y=2 and y = -2 with the order of integration being dxdy

But wouldn't that be assuming that those two planes intersect each other by equating the both of them?

Sorry about the bolding...didn't know...just used the default layout. Thanks btw. I slept over it and I was thinking to use the z=f(x,y) function as my integrating function as you said over the region bounded by the two curves. But why use between y=2 & y=-2 though? I know that I have to have...